Download Chapter 4 Oscillatory Motion

Chapter 4
Oscillatory Motion
4.1
4.1.1
The Important Stuff
Simple Harmonic Motion
In this chapter we consider systems which have a motion which repeats itself in time, that is,
it is periodic. In particular we look at systems which have some coordinate (say, x) which
has a sinusoidal dependence on time. A graph of x vs. t for this kind of motion is shown in
Fig. 4.1. Suppose a particle has a periodic, sinusoidal motion on the x axis, and its motion
takes it between x = +A and x = −A. Then the general expression for x(t) is
x(t) = A cos(ωt + φ)
(4.1)
A is called the amplitude of the motion. For reasons which will become clearer later, ω is
called the angular frequency. We say that a mass which has a motion of the type given
in Eq. 4.1 undergoes simple harmonic motion.
From 4.1 we see that when the time t increases by an amount 2π
, the argument of the
ω
cosine increases by 2π and the value of x will be the same. So the motion repeats itself
after a time interval 2π
, which we denote as T , the period of the motion. The number of
ω
x
t
Figure 4.1: Plot of x vs. t for simple harmonic motion. (t and x axes are unspecified!)
69
70
CHAPTER 4. OSCILLATORY MOTION
oscillations per time is given by f = T1 , called the frequency of the motion:
T =
2π
ω
f=
1
ω
=
T
2π
(4.2)
Rearranging we have a formula for ω in terms of f or T :
2π
T
ω = 2πf =
(4.3)
Though ω (angular frequency) and f (frequency) are closely related (with just a factor of
2π between them, we need to be careful to distinguish them; to help in this, we normally
express ω in units of rad
and f in units of cycle
, or Hz (Hertz). However, the real dimensions
s
s
of both are 1s in the SI system.
From x(t) we get the velocity of the particle:
v(t) =
dx
= −ωA sin(ωt + φ)
dt
(4.4)
a(t) =
dv
= −ω 2A cos(ωt + φ)
dt
(4.5)
and its acceleration:
We note that the maximum values of v and a are:
amax = ω 2 A
vmax = ωA
(4.6)
The maximum speed occurs in the middle of the oscillation. (The slope of x vs. t is greatest
in size when x = 0.) The magnitude of the acceleration is greatest at the ends of the
oscillation (when x = ±A).
Comparing Eq. 4.5 and Eq. 4.1 we see that
d2 x
= −ω 2 x
dt2
(4.7)
which is the same as a(t) = −ω 2x(t). Using 4.1 and 4.4 and some trig we can also arrive at
a relation between the speed |v(t)| of the mass and its coordinate x(t):
q
|v(t)| = ωA| sin(ωt + φ)| = ωA 1 − cos2 (ωt + φ)
v
u
u
t
x(t)
= ωA 1 −
A
!2
.
(4.8)
We could also arrive at this relation using energy conservation (as discussed below). Note,
if we are given x we can only give the absolute value of v since there are two possibilities for
velocity at each x (namely a ± pair).
71
4.1. THE IMPORTANT STUFF
k
x
m
Figure 4.2: Mass m is attached to horizontal spring of force constant k; it slides on a frictionless surface!
4.1.2
Mass Attached to a Spring
Suppose a mass m is attached to the end of a spring of force constant k (whose other end is
fixed) and slides on a frictionless surface. This system is illustrated in Fig. 4.2. Then if we
measure the coordinate x of the mass from the place where it would be if the spring were at
its equilibrium length, Newton’s 2nd law gives
Fx = −kx = max = m
d2 x
,
dt2
and then we have
d2 x
k
=
−
x.
dt2
m
Comparing Eqs. 4.9 and 4.7 we can identify ω 2 with
(4.9)
k
m
so that
s
k
(4.10)
m
From the angular frequency ω we can find the period T and frequency f of the motion:
ω=
2π
m
= 2π
T =
ω
k
r
1
1
f= =
T
2π
s
k
m
(4.11)
It should be noted that ω (and hence T and f) does not depend on the amplitude A
of the motion of the mass. In reality, of course if the motion of the mass is too large then
then spring will not obey Hooke’s Law so well, but as long as the oscillations are “small”
the period is the same for all amplitudes.
In the lab, it’s much easier to work with a mass bobbing up and down on a vertical
spring. One can (and should!) ask if we can still use the same formulae for T and f, or if
gravity (g) enters in somehow. In fact, the same formulae (Eq. 4.11) do apply in this case.
To be more clear about the vertical mass–spring system, we show such a system in
Fig. 4.3. In (a), the spring is oriented vertically and has some unstretched length. (We are
ignoring the mass of the spring.) When a mass m is attached to the end, the system will be
72
CHAPTER 4. OSCILLATORY MOTION
x
m
m
(b)
(a)
(c)
Figure 4.3: (a) Unstretched vertical spring of force constant k (assumed massless). (b) Mass attached to
spring is at equilibrium when the spring has been extended by a distance mg/k. (c) Mass will undergo small
oscillations about the new equilibrium position.
at equilibrium when the spring has been extended by some length y; balancing forces on the
mass, this extension is given by:
ky = mg
=⇒
y=
mg
.
k
When the mass is disturbed from its equilibrium position, it will undergo harmonic oscillations which can be described by some coordinate x, where x is measured from the new
equilibrium position of the end of the spring. Then the motion is just like that of the
horizontal spring.
Finally, we note that for more precise work with a real spring–mass system one does need
to take into account the mass of the spring. If the spring has a total mass ms , one can show
that Eq. 4.10 should be modified to:
ω=
v
u
u
t
k
m+
ms
3
(4.12)
That is, we replace the value of the mass m by m plus one–third the spring’s mass.
4.1.3
Energy and the Simple Harmonic Oscillator
For the mass–spring system, the kinetic energy is given by
K = 12 mv 2 = 21 mω 2A2 sin2 (ωt + φ)
(4.13)
U = 12 kx2 = 12 kA2 cos2(ωt + φ) .
(4.14)
and the potential energy is
Using ω 2 =
k
m
in 4.13 we then find that the total energy is
E = K + U = 21 kA2 [sin2 (ωt + φ) + cos2 (ωt + φ)]
73
4.1. THE IMPORTANT STUFF
and the trig identity sin2 θ + cos2 θ = 1 gives
E = 21 kA2
(4.15)
showing that the energy of the simple harmonic oscillator (as typified by a mass on a spring)
is constant and is equal to the potential energy of the spring when it is maximally extended
(at which time the mass is motionless).
It is useful to use the principle of energy conservation to derive some general relations for
1–dimensional harmonic motion. (We will not use the particular parameters for the mass–
spring system, just the quantities contained in Eq. 4.1, which describes the motion of a mass
m along the x axis. From Eq. 4.13 we have the kinetic energy as a function of time
K = 12 mv 2 = 21 mω 2A2 sin2 (ωt + φ)
Now the maximum value of the kinetic energy is 12 mω 2 A2, which occurs when x = 0. Since
we are free to fix the “zero–point” of the potential energy, we can agree that U(x) = 0 at
x = 0. Then the total energy of the system must be equal to the maximum (i.e. x = 0 value
of the kinetic energy:
E = 12 mω 2A2
Then using these expressions, the potential energy of the system is
U = E−K
= 21 mω 2 A2 − 21 mω 2A2 sin2(ωt + φ) = 21 mω 2 A2(1 − sin2 (ωt + φ))
= 12 mω 2 A2 cos2 (ωt + φ)
= 12 mω 2 x2
Of course, for the mass–spring system U is given by 21 kx2, which gives the relation mω 2 = k,
q
k
or ω = m
, which we’ve already found. If we use the relation vmax = ωA then the potential
energy can be written as
2
mvmax
U(x) = 12 mω 2 x2 = 12
x2
(4.16)
2
A
4.1.4
Relation to Uniform Circular Motion
There is a correspondence between simple harmonic motion and uniform circular motion,
which is illustrated in Fig. 4.4 (a) and (b). In (a) a mass point moves in a horizontal circular
path with uniform circular motion at a radius R (for example, it might be glued to the edge
of a spinning disk of radius R). Its angular velocity is ω, so its location is given by the
time–varying angle θ, where
θ(t) = ωt + φ
.
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CHAPTER 4. OSCILLATORY MOTION
y
R
q(t)
x
-R
+R
0
x
x
(b)
(a)
Figure 4.4: (a) Mass point moves in a horizontal circle of radius R. The angular velocity of its motion is
ω. A guy with a big nose (seen from above) is observing the motion of the mass at the level of the circle.
He sees only the x coordinate of the point’s motion. (b) Motion of the mass as seen by the guy with the big
nose. The projection of the motion is the same as simple harmonic motion with angular frequency ω and
amplitude R.
I, M
Pivot
q
d
q
L
CM
m
(a)
(b)
Figure 4.5: (a) Simple pendulum. (b) Physical pendulum.
In 4.4 (b) we show the motion of the mass as it would be seen by someone looking
toward the +y direction at the level of the disk. Such an observer sees only the changing x
coordinate of the mass’s motion. Since x = R cos θ, the observed coordinate is
x(t) = R cos(θ(t)) = R cos(ωt + φ) ,
the same as Eq. 4.1. The motion of the corresponding (projected) harmonic oscillator has
an angular frequency of ω and an amplitude of R.
4.1.5
The Pendulum
We start with the simple pendulum, which has just a small mass m hanging from a string
of length L whose mass we can ignore. (See Fig. 4.5 (a).) The mass is set into motion so
75
4.1. THE IMPORTANT STUFF
that it moves in a vertical plane. One can show that if θ is the angle which the string makes
with the vertical, it obeys the differential equation:
d2 θ
g
=
−
sin θ
dt2
L
One should note that this is not of the form given in Eq. 4.7.
Things are much simpler when we restrict θ to be “small” at all times. If that is the case,
then we can use the approximation sin θ ≈ θ, which is true if we are measuring θ in radians.
Then the differential equation becomes
g
d2 θ
=− θ
(4.17)
2
dt
L
Comparison of this equation with Eq. 4.7 lets us identify the angular frequency of the
motion:
r
g
ω=
(4.18)
L
s
L
2π
ω
1 g
T =
= 2π
f=
=
(4.19)
ω
g
2π
2π L
The (perhaps) surprising thing about Eqs. 4.18 and 4.19 is that they have no dependence
on the mass suspended from the string or on the amplitude of the swing. . . as long as it is a
small angle!
r
θ(t) = θmax cos(ωt + φ)
(4.20)
We must always keep our assumption of “small” θ in the back of our minds whenever we
do a problem with a pendulum. The formulae giving T and f become less accurate as θmax
gets too big.
An important generalization of the simple pendulum is that of a rigid body which is free
to rotate in a plane about some (frictionless!) pivot. Such a system is known as a physical
pendulum and is diagrammed in Fig. 4.5 (b).
Suppose we look at the line which joins the pivot to the center of mass of the object. If
θ is the angle which this line makes with the vertical, and if we again use the approximation
sin θ ≈ θ, one can show that it obeys the differential equation
d2 θ
Mgd
=−
θ
(4.21)
2
dt
I
where d is the distance between the pivot and the center of mass, M is the mass of the
object and I is the moment of inertia of the object about the given axis. (Note: the axis is
probably not at the center of mass; if it were, the mass wouldn’t oscillate!)
Following the usual procedure we find the period T :
s
2π
I
T =
= 2π
ω
Mgd
(4.22)
76
4.2
4.2.1
CHAPTER 4. OSCILLATORY MOTION
Worked Examples
Simple Harmonic Motion
1. The displacement of a particle at t = 0.25 s is given by the expression x =
(4.0 m) cos(3.0πt + π) where x is in meters and t is in seconds. Determine (a) the
frequency and period of the motion, (b) the amplitude of the motion, (c) the
phase constant, and (d) the displacement of the particle at t = 0.25 s. [Ser4 13-1]
(a) We compare the given function x(t) with the standard form for simple harmonic motion
given in Eq. 4.1. This gives us the angular frequency ω:
ω = 3.0π
rad
s
and from this we can get the frequency and period:
3.0π rad
ω
s
f=
=
= 1.50 Hz
2π
2π
T =
1
1
=
= 0.667 s
f
(1.50 s−1 )
(b) We easily read off the amplitude as the factor (a length) which multiplies the cosine
function:
A = 4.0 m
(c) Again, comparison with Eq. 4.1 gives
φ=π
(d) At t = 0.25 s the displacement (i.e. the coordinate) of the particle is:
x(0.25 s) = (4.0 m) cos((3.0π)(0.25) + π) = (4.0 m) cos((1.75)π)
= (4.0 m)(0.707) = 2.83 m
2. A loudspeaker produces a musical sound by means of the oscillation of a
diaphragm. If the amplitude of oscillation is limited to 1.0 × 10−3 mm, what frequencies will result in the magnitude of the diaphragm’s acceleration exceeding
g? [HRW5 16-5]
We are given the amplitude of the diaphragm’s motion, A = 1.0×10−3 mm = 1.0×10−6 m.
From Eq. 4.6, the maximum value of the acceleration is amax = Aω 2. So then the angular
frequency that results in a maximum acceleration of g is
ω2 =
(9.8 sm2 )
amax
=
= 9.8 × 106 s−2
A
(1.0 × 10−6 m)
77
4.2. WORKED EXAMPLES
ω = 3.1 × 103 s−1 .
=⇒
This corresponds to a frequency of
f=
ω
(3.1 × 103 s−1 )
=
= 5.0 × 102 Hz
2π
2π
At frequencies larger than 500 Hz, the acceleration of the diaphragm will exceed g.
3. The scale of a spring balance that reads from 0 to 15.0 kg is 12.0 cm long.
A package suspended from the balance is found to oscillate vertically with a
frequency of 2.00 Hz. (a) What is the spring constant? (b) How much does the
package weigh? [HRW5 16-6]
(a) The data in the problem tells us us that the spring within the balance increases in length
by 12.0 cm when a weight of
W = mg = (15.0 kg)(9.80 sm2 ) = 147 N
is pulls downward on its end. So the force constant of the spring must be
k=
F
(147 F)
N
=
= 1225 m
x
(12 × 10−2 m)
(b) Eq. 4.11 we have the frequency of oscillation of the mass–spring system in terms of the
spring constant and the attached mass. We have the frequency and spring constant and we
can solve to get the mass of the package:
1
f=
2π
s
Plug in the numbers:
m=
k
m
=⇒
m=
k
4π 2 f 2
N
(1225 m
)
= 7.76 kg
2
−1
4π (2.00 s )2
That’s the mass of the package; its weight is
W = mg = (7.76 kg)(9.80 sm2 ) = 76 N
4. In an electric shaver, the blade moves back and forth over a distance of 2.0 mm
in simple harmonic motion, with frequency 120 Hz. Find (a) the amplitude, (b)
the maximum blade speed, and (c) the magnitude of the maximum acceleration.
[HRW5 16-9]
(a) The problem states that the full distance that the blade travels on each back-and-forth
swing is 2.0 mm, but the full swing distance is twice the amplitude. So the amplitude of the
motion is A = 1.0 mm.
78
CHAPTER 4. OSCILLATORY MOTION
(b) From Eq. 4.6 we have the maximum speed of an oscillating mass in terms of the amplitude
and frequency:
vmax = ωA = 2πfA = 2π(120 s−1 )(1.0 × 10−3 m) = 0.75 ms
(c) From Eq. 4.6 we also have magnitude of the maximum acceleration of an oscillating mass
in terms of the amplitude and frequency:
amax = ω 2 A = (2πf)2 A = 4π 2 (120 s−1 )2 (1.0 × 10−3 m) = 570 sm2
5. The end of one of the prongs of a tuning fork that executes simple harmonic
motion of frequency 1000 Hz has an amplitude of 0.40 mm. Find (a) the maximum
acceleration and (b) the maximum speed of the end of the prong. Find (c)
the acceleration and (d) the speed of the end of the prong when the end has a
displacement of 0.20 mm [HWR5 16-22]
(a) Since we have the amplitude A of the prong’s motion, and we can easily find the angular
frequency ω:
ω = 2πf = 2π(1000 Hz) = 6.28 × 103 s−1
we can use Eq. 4.6 to find the maximum value of a:
amax = ω 2A = (6.28 × 103 s−1 )2 (0.400 × 10−3 m)
= 1.6 × 104 sm2
(b) Likewise, from the same equation we find the maximum speed of the prong’s tip:
vmax = ωA = (6.28 × 103 s−1 )(0.400 × 10−3 m)
= 2.5 ms
(c) Equation 4.7 relates the acceleration a and coordinate x at all times. When the displacement of the prong’s tip is 0.20 mm (half of its maximum) we find
a = −ω 2 x = −(6.28 × 103 s−1 )2(0.20 × 10−3 m) = −7.9 × 103
m
s2
(d) We have already given a relation between |v| (speed) and x in Eq. 4.8. We use it here
to find the speed when x = 0.20 mm:
v
u
u
t
x(t)
|v| = ωA 1 −
A
3 −1
!2
= (6.28 × 10 s )(0.40 × 10
= 2.2 ms
−3
s
0.20 mm
m) 1 −
0.40 mm
2
79
4.2. WORKED EXAMPLES
k
m
Frictionless
M
Figure 4.6: Mass M is attached to a spring and oscillates on a frictionless surface. Another block of mass
m is on top!
4.2.2
Mass Attached to a Spring
6. A 7.00 − kg mass is hung from the bottom end of a vertical spring fastened to
an overhead beam. The mass is set into vertical oscillations having a period of
2.60 s. Find the force constant of the spring. [Ser4 13-11]
The formulae in Eq. 4.11 hold even if the mass–spring system oscillates vertically (just
as long as we can neglect the mass of the spring). Then we can solve for the force constant:
T = 2π
r
m
k
=⇒
T2 =
4π 2 m
k
=⇒
k=
4π 2 m
T2
and the numbers give us
k=
4π 2(7.00 kg)
N
= 40.9 kg
= 40.9 m
.
s2
(2.60 s)2
N
.
The force constant of the spring is 40.9 m
N
7. Two blocks (m = 1.0 kg and M = 10 kg) and a spring (k = 200 m
) are arranged
on a horizontal, frictionless surface as shown in Fig. 4.6. The coefficient of static
friction between the two blocks is 0.40. What is the maximum possible amplitude
of simple harmonic motion of the spring–block system if no slippage is to occur
between the blocks? [HRW5 16-25]
We first look at what happens when the two blocks oscillate together. In that case it is
legal to regard the mass on the spring as a single mass whose value is M = M + m = 11.0 kg.
We know the spring constant, so using Eq. 4.10 the angular frequency of the motion is
ω=
s
k
=
M
v
u
u
t
N
200 m
= 4.26 s−1
11.0 kg
80
CHAPTER 4. OSCILLATORY MOTION
N
fs
fs
m
mg
Figure 4.7: The forces acting on mass m in Example 7. (The force of static friction changes direction and
magnitude during the motion of mass m.)
During the motion, the large mass oscillates with this frequency and so does the small mass
since they move together. But note, the spring is attached only to the large mass; what is
making the small mass move back and forth? The answer is static friction.
We make a diagram of the forces which act on the small mass. This is shown in Fig. 4.7.
We have the force of gravity mg pointing down, the normal force N from the big block
pointing up and also the force of static friction fs , which can point either to the right or
to the left, depending on the current position of m during the oscillation! The magnitude
and direction of the static friction force fs are not constant; the value of fs depends on the
acceleration of the co-moving blocks (assuming there is no slipping so that they are indeed
co-moving).
There is no vertical motion of the small block so clearly
N = mg = (1.00 kg)(9.80 sm2 ) = 9.80 N .
But having the normal force (between the surfaces of the two blocks) we know the maximum
possible magnitude of the static friction force, namely:
fsmax = µs FN = µs mg
and since that is the only sideways force on mass m, from Newton’s 2nd Law, the maximum
possible magnitude of its acceleration — assuming no slipping — is
ano−slip
=
max
fsmax
µs mg
=
= µs g .
m
m
Now, if the two blocks are moving together and oscillating with amplitude A, then the
maximum value of the acceleration is given by Eq. 4.6, namely amax = ω 2 A, which of course
will get larger if A gets larger. By equating this maximum acceleration of the motion to the
value we just found, we arrive at a condition on the maximum amplitude A such that no
slipping will occur:
ano−slip
= ω 2 Amax
max
=⇒
µs g = ω 2 Amax
81
4.2. WORKED EXAMPLES
which gives:
Amax =
4.2.3
(0.40)(9.80 sm2 )
µs g
=
= 0.216 m
ω2
(4.26 s−1 )2
Energy and the Simple Harmonic Oscillator
8. A particle executes simple harmonic motion with an amplitude of 3.00 cm. At
what displacement from the midpoint of its motion does its speed equal one half
of its maximum speed? [Ser4 13-23]
The maximum speed occurs in the center of the motion, where there is no potential
energy. So the total energy is given by
2
E = 21 mvmax
At the point(s) where v = 21 vmax the potential energy is not zero; rather it is given by
U = E − 21 mv 2
2
= 12 mvmax
− 21 mv 2
vmax 2
2
1
1
= 2 mvmax − 2 m
2
1 1
2
2
=
−
mvmax
= 83 mvmax
2 8
But we also have from Eq. 4.16 the result
U(x) =
1
2
2
mvmax
x2
A2
And combining these expressions gives the corresponding value of x:
2
1 mvmax 2
x
2
2
A
2
= 38 mvmax
Solve for x:
3
x = A2
4
2
=⇒
√
√
3
3
x=±
A=±
(3.00 cm) = ±2.60 cm
2
2
The mass has half its maximum speed at x = ±2.60 cm.
The problem can also be worked just using Eqs. 4.1 and 4.4. The problem gives no data
about any specific value of t so we are free to choose φ = 0 for simplicity. Then
x(t) = A cos(ωt)
and
v(t) = −ωA sin(ωt) = −vmax sin(ωt)
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CHAPTER 4. OSCILLATORY MOTION
and for the times t at which the speed of the mass is half the maximum value, we must have
the condition
sin(ωt) = ± 12 .
But when this is true we have
cos2 (ωt) = 1 − sin2 (ωt) = 1 −
or
1
4
=
3
4
√
3
cos(ωt) = ±
2
and that gives
4.2.4
√
√
3
3
x = ±A
= ±(3.00 cm)
= ±2.60 cm
2
2
The Simple Pendulum
9. A simple pendulum has a period of 2.50 s. (a) What is its length? (b) What
would its period be on the Moon, where gMoon = 1.67 sm2 ? [Ser4 13-25]
(a) Using Eq. 4.19 we solve for the length:
T = 2π
s
L
g
T 2 = 4π 2
=⇒
L
g
=⇒
L=
T 2g
4π 2
and the numbers give:
(2.50 s)2 (9.80 sm2 )
= 1.55 m
4π 2
The length of the pendulum is 1.55 m.
L=
(b) If we take this pendulum to the Moon, its length will be the same, but the acceleration
of gravity will be different. Using the new value of g in Eq. 4.19 we find
TMoon = 2π
s
L
gMoon
=
v
u
u (1.55 m)
2π t
(1.67 sm2 )
= 6.06 s
The pendulum’s period on the Moon is 6.06 s.
10. If a simple pendulum with length 1.50 m makes 72.0 oscillations in 180 s, what
is the acceleration of gravity at its location? [HRW5 16-59]
We find the frequency f of this pendulum:
f=
72.0
= 0.400 Hz
180 s
83
4.2. WORKED EXAMPLES
qmax
L cos qmax
L
2.06 m/s
(b)
(a)
Figure 4.8: (a) Pendulum starts with speed 2.06 ms at the bottom of the swing. (b) It attains a maximum
angular displacement θmax .
Then from Eq. 4.19 we can solve for the value of g:
1
f=
2π
r
g
L
=⇒
(2πf)2 =
g
L
=⇒
g = 4π 2 f 2 L
Plug in the numbers:
g = 4π 2 (0.400 s−1 )2(1.50 m) = 9.47 sm2
The acceleration of gravity at this location is 9.47 sm2 .
11. A simple pendulum having a length of 2.23 m and a mass of 6.74 kg is given
an initial speed of 2.06 ms at its equilibrium position. Assume it undergoes simple harmonic motion and determine its (a) period, (b) total energy, and (c)
maximum angular displacement. [Ser4 13-59]
The problem is diagrammed in Fig. 4.8 (a).
I will answer the parts of this question in a different order; one reason for this is that
part (c) (maximum value of θ) can clearly be found using energy conservation. Finding the
maximum angular displacement will then give the period.
First off, if we measure height from the bottom of the pendulum’s swing, then in its
initial position it has no potential energy but a kinetic energy equal to
K = 12 mv 2 = 12 (6.74 kg)(2.06 ms )2 = 14.3 J
so the total energy of the system is 14.3 J.
Now when the mass reaches its maximum angular displacement (say, θmax) it is at a
height
ymax = L − L cos θmax = L(1 − cos θmax) .
At that time all of the energy of the particle is potential energy and using energy conservation, we can solve for θmax :
E = mgymax = mgL(1 − cos θmax) = 14.3 J
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CHAPTER 4. OSCILLATORY MOTION
(1 − cos θmax) =
(14.3 J)
(14.3 J)
=
= 9.71 × 10−2
mgL
(6.74 kg(9.80 sm2 )(2.23 m)
cos θmax = 1 − 9.71 × 10−2 = 9.03 × 10−1
θmax = 25.4◦ = 0.444 rad
This is the exact answer for θmax . Now, one might wonder if 25.4◦ is small enough so that
our calculation of the period of the motion is very accurate, but we forge on anyway!
Now, we are given the linear speed at the bottom of the swing, but the pendulum’s
(harmonic) motion has to do with its angle. We need to relate the two.
From Eq. 1.10 we can get the angular velocity of the mass at the bottom of the swing:
dθ
dt
!
=
0
(2.06 ms )
v0
=
= 0.924 rad
s
L
(2.23 m)
But from 4.20 we have θ(t) = θmax cos(ωt + φ) so that the angular velocity of the pendulum
at all times is
dθ
= −ωθmax cos(ωt + φ)
dt
so that the maximum angular speed (namely at the bottom of the swing) is
dθ
dt
!
= ωθmax
max
found above. So we can get ω:
and this is the same as the 0.924 rad
s
dθ
dt
!
= 0.924
rad
s
= ωθmax
=⇒
max
(0.924 rad
)
s
ω=
= 2.08 rad
s
(0.444 rad)
(It is true that the units don’t look right on that last one, but keep in mind that “radian”
is really dimensionless.)
Having ω, the angular frequency of the pendulum’s oscillations, we go on to get the
period:
2π
= 3.02 s
T =
ω
Summing up what problem asked for, we have:
(a) T = 3.02 s
4.2.5
(b) E = 14.3 J
(c) θmax = 0.444 rad
Physical Pendulums
12. A physical pendulum in the form of a planar body moves in simple harmonic
motion with a frequency of 0.450 Hz. If the pendulum has a mass of 2.20 kg and
the pivot is located 0.350 m from the center of mass, determine the moment of
inertia of the pendulum. [Ser4 13-33]
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4.2. WORKED EXAMPLES
From Eq. 4.22 we have an expression for the frequency of a “physical pendulum”:
1
1
f= =
T
2π
s
Mgd
I
We have all the values that we need to solve for I:
f2 =
Mgd
4π 2I
=⇒
I=
Mgd
4π 2 f 2
Plug in the numbers:
I=
(2.20 kg)(9.80 sm2 )(0.350 m)
= 0.944 kg · m2
4π 2(0.450 s−1 )2
13. A thin disk of mass 5 kg and radius 20 cm is suspended by a horizontal axis
perpendicular to the disk through the rim. The disk is displaced slightly from
equilibrium and released. Find the period of the subsequent simple harmonic
motion. [Tip4 14-57]
We need to find the moment of inertia of the disk when it rotates around an axis at the
rim of the disk. For this, we use the Parallel Axis Theorem of Chapter 1. With ICM given
by ICM = 21 MR2 and recognizing that in this problem, the axis has been shifted a distance
D = R away from the center of mass, we find:
I = ICM + MD2 = 12 MR2 + MR2 = 23 MR2
Then we can find the period of the oscillatory motion from Eq. 4.22. Note, the distance of
the axis from the center of mass (called D in that formula) is R:
s
I
T = 2π
= 2π
Mgd
v
u3
u MR2
t2
s
v
u
u 3(0.20 m)
3R
= 2π
= 2π t
= 1.1 s
MgR
2g
2(9.80 sm2 )
The period of this pendulum is 1.1 s. Interestingly enough, the answer did not depend on
the mass of the disk.
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CHAPTER 4. OSCILLATORY MOTION